Question

Prove that 7√5 is irrational?
No copied answer ❌.

Answer 1

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Answer 2

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$Let \: us \: assume \: that \: the \: 7 \sqrt{5} \: is \: rational.$

So, that we can find coprime integers a and b

( ≠ 0 ) such that

$7 \sqrt{5 } = \frac{a}{b}$

$\sqrt{5} = \frac{a}{7b}$

Since, a and b are integers.

$\frac{a}{7b} \: is \: rational \: and \: so \: \sqrt{5} \: is \: rational.$

$But \: this \: contradicts \: the \: fact \: that \: \sqrt{5} \: is \: irrational.$

$So \: we \: conclude \: that \: 7 \sqrt{5} \: is \: irrational.$